Question

7) Suppose you wanted to test to see if the population mean of cotinine levels in the 40 smokers from question 6 (SEE BELOW FOR QUESTION 6) is equal to 225 ng/ml. Using the confidence interval method what would your decision be using: (3 pts)

a) A 90% confidence interval estimate of the mean cotinine level of all smokers.

b) A 98% confidence interval estimate of the mean cotinine level of all smokers.

c) A 92% confidence interval estimate of the mean cotinine level of all smokers.

QUESTION 6) When people smoke, the nicotine they absorb is converted to cotinine, which can be measured. A sample of 40 smokers has a mean cotinine level of 172.5 ng/ml. Assume the σ is known to be 119.5 ng/ml.

Answer 1

Let's write the given information.

n = sample size = 40

sample mean = = 1725

standard deviation = = 1195

a) Let's use minitab:

Step 1) Click on Stat>>>Basic Statistics >>1 sample Z...

Click on summarized data

Sample size = 40

Mean = 172.5

Standard deviation = 119.5

Step 2) then click on Option

Confidence level = 90.0

Alternative "not equal "

Look the following image:

then click on OK

again click on OK

So we get the following output:

From the above output the 90% confidence interval for population mean is  (141.4, 203.6)

Since 225 is not included in the above interval , so we reject the null hypothesis and support the alternative hypothesis.

b) Here we need to plug 98 insted of 90 and all the above procedure is same.

Look the following output.

From the above output the 98% confidence interval for population mean is  (128.5, 216.5)

Since 225 is not included in the above interval , so we reject the null hypothesis and support the alternative hypothesis.

c) Here Confidence level : 92 and all the procedure in part a) is same.

So we get the following output:

From the above output the 92% confidence interval for population mean is (139.4, 205.6)

Since 225 is not included in the above interval , so we reject the null hypothesis and support the alternative hypothesis.